integral of (x^2)/(x^2-6x+25)

解

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x + C

解答ステップ

\int \frac{x ^{2}}{x ^{2} - 6 x + 25} d x

\frac{x ^{2}}{x ^{2} - 6 x + 25} = \frac{6 x - 25}{x ^{2} - 6 x + 25} + 1

= \int \frac{6 x - 25}{x ^{2} - 6 x + 25} + 1 d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{6 x - 25}{x ^{2} - 6 x + 25} d x + \int 1 d x

\int \frac{6 x - 25}{x ^{2} - 6 x + 25} d x = \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3)))

\int 1 d x = x

= \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x

定数を解答に追加する

= \frac{1}{4} (12 ln \frac{1}{16} (x - 3)^{2} + 1 - 7 arctan (\frac{1}{4} (x - 3))) + x + C